What is the integral of (18x^2)/(\sqrt[3]{x^3+4)} ?

The integral of (18x^2)/(\sqrt[3]{x^3+4)} is 9(x^3+4)^{2/3}+C

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integral of (18x^2)/(\sqrt[3]{x^3+4)}

\int \frac{18 x ^{2}}{3 x ^{3} + 4} d x

9 (x^{3} + 4)^{\frac{2}{3}} + C

Solution steps

\int \frac{18 x ^{2}}{3 x ^{3} + 4} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 18 \cdot \int \frac{x ^{2}}{3 x ^{3} + 4} d x

Apply u-substitution

: \int \frac{1}{3 3 u} d u

= 18 \cdot \int \frac{1}{3 3 u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 18 \cdot \frac{1}{3} \cdot \int \frac{1}{3 u} d u

Apply the Power Rule

: \frac{3}{2} u^{\frac{2}{3}}

= 18 \cdot \frac{1}{3} \cdot \frac{3}{2} u^{\frac{2}{3}}

Substitute back

u = x^{3} + 4

= 18 \cdot \frac{1}{3} \cdot \frac{3}{2} (x^{3} + 4)^{\frac{2}{3}}

Simplify

18 \cdot \frac{1}{3} \cdot \frac{3}{2} (x^{3} + 4)^{\frac{2}{3}} : 9 (x^{3} + 4)^{\frac{2}{3}}

= 9 (x^{3} + 4)^{\frac{2}{3}}

Add a constant to the solution

= 9 (x^{3} + 4)^{\frac{2}{3}} + C

\int x^{\frac{1}{3}} + 4 d x

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What is the integral of (18x^2)/(\sqrt[3]{x^3+4)} ?
The integral of (18x^2)/(\sqrt[3]{x^3+4)} is 9(x^3+4)^{2/3}+C